The monthly earnings of a group of business students are are normally distributed with a standard deviation of 589 dollars. a researcher wants to estimate the mean monthly earnings of all business students. find the sample size needed to have a confidence level of 95% and a margin of error of 132 dollars.

Margin of error = +/- Z*SD/sqrt (n)

Where, Z = 1.96 at 95% confidence interval, SD =standard deviation = 589 dollars, n= sample size (to be calculated)

For  the values given;

132= 1.96*589/Sqrt (n) => Sqrt (n) = 1.96*589/132 = 8.7458 => n= 8.7458^2 = 76.49

Rounding up to near a person, n (sample size)= 77 students